# 2022考研数二第04题解析：二元偏导数、变上限积分求导

## 二、解析

\begin{aligned} F ( x , y ) \\ \\ & = \int _{ 0 } ^ { x – y } ( x – y – t ) f ( t ) \mathrm{ ~d } t \\ \\ & = ( x – y ) \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t – \int _{ 0 } ^ { x – y } t f ( t ) \mathrm{ ~d } t \\ \\ & = x \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t – y \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t – \int _{ 0 } ^ { x – y } t f ( t ) \mathrm{ ~d } t \end{aligned}

\begin{aligned} \frac { \partial F } { \partial x } \\ \\ & = \left[ \textcolor{springgreen}{x} \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} – \textcolor{yellow}{y \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} – \textcolor{yellow}{\int _{ 0 } ^ { x – y } t f ( t ) \mathrm{ ~d } t} \right]^{\prime}_{\textcolor{springgreen}{x}} \\ \\ & = \left[ \textcolor{springgreen}{x} \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} \right] ^{\prime}_{x} \\ \\ & = \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} \end{aligned}

\begin{aligned} \frac { \partial F } { \partial y } \\ \\ & = \left[ \textcolor{yellow}{x \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} – \textcolor{springgreen}{y} \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} – \textcolor{yellow}{\int _{ 0 } ^ { x – y } t f ( t ) \mathrm{ ~d } t} \right] ^{\prime}_{\textcolor{springgreen}{y}} \\ \\ & = \left[ – \textcolor{springgreen}{y} \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} \right] ^{\prime} _{\textcolor{springgreen}{y}} \\ \\ & = \textcolor{yellow}{- \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} \end{aligned}

$$\textcolor{magenta}{ \boldsymbol{ \frac { \partial F } { \partial x } = – \frac { \partial F } { \partial y } } }$$

\begin{aligned} \frac { \partial ^ { 2 } F } { \partial x ^ { 2 } } \\ \\ & = \frac{\partial}{\partial x} \left( \frac { \partial F } { \partial x } \right) \\ \\ & = \left( \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t \right) ^{\prime} _{x} \\ \\ & = \textcolor{orangered}{x ^{\prime}} + f(x – y) \\ \\ & = f ( x – y ) \end{aligned}

\begin{aligned} \frac { \partial ^ { 2 } F } { \partial y ^ { 2 } } \\ \\ & = \frac{\partial}{\partial y} \left( \frac { \partial F } { \partial y } \right) \\ \\ & = \left( – \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t \right) ^{\prime} _{y} \\ \\ & = \textcolor{orangered}{-(-y) ^{\prime}} + f(x – y) \\ \\ & = f ( x – y ) \end{aligned}

$$\textcolor{magenta}{ \boldsymbol{ \frac { \partial ^ { 2 } F } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } F } { \partial y ^ { 2 } } } }$$